Answer:
50.135% (3 d.p.)
Step-by-step explanation:
To solve this problem, we can use the properties of the standard normal distribution since the scores are assumed to follow a nearly symmetric / bell-curve distribution.
The final exam scores of students in a Math class (X) are normally distributed with a mean (μ) of 152 and a standard deviation (σ) of 4.6. Therefore:
[tex]X\sim \text{N}(\mu, \sigma^2) \implies X\sim \text{N}(152, 4.6^2)[/tex]
To calculate the percent of students that score below 138.2 or above 152, we need to find:
[tex]\text{P}(X\leq 138.2)+\text{P}(X\geq 152)[/tex]
To find P(X ≤ 138.2), enter the following into a calculator's "normal cumulative distribution function (cdf)":
- Lower bound: x = -100
- Upper bound: x = 138.2
- σ = 4.6
- μ = 152
This gives the probability that a student scores 138.2 or less as:
[tex]\text{P}(X\leq 138.2)=0.00134989813...[/tex]
[tex]\text{P}(X\leq 138.2)=0.135\%[/tex]
As the mean is 152, the percent of students above the mean will be exactly 50%, so:
[tex]P(X \geq 152) = 50\%[/tex]
Therefore, the percent of students that will score below 138.2 or above 152 is:
[tex]\text{P}(X\leq 138.2)+\text{P}(X\geq 152)=0.135\%+50\%[/tex]
[tex]\text{P}(X\leq 138.2)+\text{P}(X\geq 152)=50.135\%[/tex]
So, approximately 50.135% of students will score below 138.2 or above 152.
